Let a⃗=î+2 ĵ+3 k̂ and b⃗=î+ĵ-k̂. If c⃗ is a vector such that a⃗ · c⃗=11, b⃗ ·(a⃗ × c⃗)=27 and b⃗ · c⃗=-√3|b⃗|, then |a⃗ × c⃗|² is equal to _________.

Solution

Maths sir · 4 steps

a⃗ ⊥ b⃗, so write c⃗ using a⃗, b⃗ and a⃗ × b⃗ as a basis.

1. Spot perpendicularity. a⃗ · b⃗ = 1 + 2 - 3 = 0, so a⃗, b⃗ and a⃗ × b⃗ = -5î + 4ĵ - k̂ are mutually perpendicular. Here |a⃗|² = 14, |b⃗|² = 3 and |a⃗ × b⃗|² = 42.

   c⃗ = α a⃗ + β b⃗ + γ (a⃗ × b⃗)

2. Find the coefficients. Dot with a⃗: 14α = 11, so α = 11/14. Dot with b⃗: 3β = -√3 × √3 = -3, so β = -1. Also b⃗ · (a⃗ × c⃗) = -c⃗ · (a⃗ × b⃗) = -42γ = 27, so γ = -9/14.
3. Length of c. |c⃗|² = 14α² + 3β² + 42γ² = 121/14 + 3 + 243/14 = 26 + 3 = 29.
4. So. By Lagrange's identity, |a⃗ × c⃗|² = |a⃗|²|c⃗|² - (a⃗ · c⃗)² = 14 × 29 - 121 = 285.

The common slip: 257. Taking b⃗ · c⃗ = -√3 instead of -√3|b⃗| = -3 gives |c⃗|² = 27 and the answer 257, not 285.

The idea to keep. When two given vectors are perpendicular, use them and their cross product as a basis. Each dot condition then fixes one coefficient, and Lagrange's identity finishes the job.

Answer from the NTA final answer key. Solution by PrepWiser.